2009 Main Line Layout

[Hort]

hi,
When carrying out headway calcs for a non-stopping train following a stopper, could someone explain how the additional time equates to the Textra = Tdwell + 0.5(Braking time + acceleration time)?

How is the 0.5(Braking time + acceleration time) portion derived? I had a go at deriving it using Newton's laws and substitiution but can't get my head around it!

[reuben]

Hi

You can prove it using the equations of motion - the key trick being to eliminate the distance taken to accelerate/ decelerate, which you don't know or want to know!

It's easier to do it in words:

Under constant acceleration, the mean velocity during a period of accn from 0 to V or decn from V to 0 will be V/2.

Therefore, it will take exactly twice as long to cover the same distance, during braking or acceleration, compared to going at full speed.

from equations of motion, time to accelerate = v/a, hence extra time taken to accelerate = v/2a.

This effect is repeated for both the deceleration and acceleration phases, hence Textra = 0.5 * (decn time + accn time) + Tdwell

You can also demonstrate this using a velocity-time graph, the area under this graph is the distance travelled. Comparing graphs for constant speed and station stopping, the difference in area is

V * (0.5 * (decn time + accn time) + Tdwell)

and hence 0.5 * (decn time + accn time) + Tdwell is extra time taken to cover the same distance.

hope this helps

Reuben

[Hort]

Thanks for your help